How do you differentiate #y=sqrt(x)/(1+sqrt(x))#?

Question

6613 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-29T13:18:34

#1/(2sqrtx(1+sqrtx)^2)#.

#y=sqrtx/(1+sqrtx)=(1+sqrtx-1)/(1+sqrtx)=(1+sqrtx)/(1+sqrtx)-1/(1+sqrtx)#.

#:. y=1-1/(1+sqrtx)#

Knowing that, #d/dt(1/t)=-1/t^2#, and, using Chain Rule, we have,

#dy/dx=d/dx(1)-d/dx(1/(1+sqrtx))#

#=0-{-1/((1+sqrtx)^2)}*d/dx((1+sqrtx))#

#=1/((1+sqrtx)^2)(1/2sqrtx)#

#=1/(2sqrtx(1+sqrtx)^2)#.